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Value of g - CGS, FPS, Earth, FAQs

Value of g - CGS, FPS, Earth, FAQs

Edited By Team Careers360 | Updated on Jul 02, 2025 05:00 PM IST

What is acceleration due to gravity ‘g’?

According to Newton’s second law of motion, if a force F acts on a body of mass ‘m’, it produces an acceleration ‘a’ i.e. F=ma.

If this force is due to gravity of earth, then this acceleration produced is called acceleration due to gravity denoted by ‘g’.

F=mg (equation-1)

The value of g on earth changes throughout the globe as it depends on factors like altitude, depth, radius of earth, etc. We will find out the reason behind this in this article later. We will also learn how to calculate the value of g (acceleration due to gravity value )when height or depth of a body w.r.t the surface of the earth changes. By equation-1 we can define g as the force of gravity acting on a body of mass m placed near or on the surface of the earth (i.e. at a nearby height or depth or on the surface of the earth).

This Story also Contains
  1. What is acceleration due to gravity ‘g’?
  2. Direction and Unit of ‘g’
  3. Relation Between g and G
  4. Radius, mass and density of earth
  5. Variation of g/factors affecting value of g
  6. Thus at centre of earth value of g is zero
Value of g - CGS, FPS, Earth, FAQs
Value of g - CGS, FPS, Earth, FAQs

Also check-

Background wave

Since g=F/m, taking m=1 we get- g=F

g is not a universal constant like G. The value of g varies from planet to planet and therefore the weight of the body which is F=mg is different in different planets. However at the surface of the earth value of g is taken to be 9.8m/s2. Hence if a body is freely falling under the influence of earth’s gravity, the acceleration with which it falls is ‘g’.

The value of g is independent of shape, mass or size of a body. It only depends on the mass of earth and radius of earth (or any planet). Thus if 2 bodies of different masses are allowed to fall from the same height h, it reaches the earth’s surface simultaneously. Look at the picture below, Galileo was the first person to do this experiment by dropping 2 spheres of different masses from the top of Leaning Tower of Pisa and established the fact that both of them reached the surface simultaneously.

Experiment of dropping of 2 spheres of different masses by Galileo

However if 2 bodies have the same masses but different volumes, they will not reach the earth’s surface at the same time because the larger the volume of the body, the larger is the upthrust from air. Hence the body with larger volume will take more time than the body with lower volume to reach the earth’s surface.

Direction and Unit of ‘g’

Acceleration due to gravity ‘g’ is a vector quantity and is always directed towards the centre of the earth. If we consider the freely falling body as the origin in Cartesian coordinates, the direction of g is always towards –Y axis. The S.I. unit of g is m/s2 or N/kg. The dimensional formula for g is M0L1T-2.Value of g in fps (feet per second squared) is 32.1741 feet per second squared.

Relation Between g and G

Let us consider earth to be a sphere of mass M, radius R with the centre of earth as O. Now, if a body of mass m is placed on the surface of earth, g is the acceleration due to gravity acting on it.

A body of mass m is placed on the surface the earth. (Fig-1)

See the above figure, the force on body of mass m due to earth (mass taken to be M) is-

F=GMm/R2 => mg= GMm/R2 =>g=GM/R2 (equation-2)

From the above equation we can see that the value of g doesn’t depend on mass, size, shape of the body but on mass and radius of the earth.

Radius, mass and density of earth

Now you must be thinking ‘what is the radius of the earth?’ and ‘what is the mass of the earth?’ and how to calculate them.From equation-2 above, we can calculate the mass of the earth. We know that the radius of earth is R=6400km. By equation 2-

M=gR2 /G , where g= 9.8m/s2 and G=6.6710-11Nm2kg-2 - (equation-3)

( unit of gravitation constant in cgs unit is measured in centimeter cube per gram per second square)

Putting the above values, we get M=6.0181024kg

Thus mass of earth in kg is- M=6.0181024kg

Now since we are considering earth to be a spherical body of radius R, then densityρ=MassVolume=M/(43.πR3)=3M/4R3

Then putting equation-3 in equation-2-

ρ=3g4πRG

Putting values of g, G and R, we get- ρ=5.5103kgm-3

Similarly, we can similarly calculate the value of g on mars or value of g on jupiter, if we know the mass and radius of mars and vice versa.

Variation of g/factors affecting value of g

The value of acceleration due to gravity g changes with height, depth and shape of the earth. Let us consider each case and find the formula for g.

a)Effect of altitude(height)- We consider earth to be a sphere of mass M and radius R having centre at O. Let acceleration due to gravity at P (on the surface of earth) be g.

Then, g=GM/R2 (equation-4)

Here, we calculate how g changes when the body of mass m is placed at a height of 'h'. (Fig-2)

If acceleration due to gravity at point Q is taken g’, at a height h above the surface of the earth, then-

g’=GM/(R+h)2 (equation-5)

Dividing equation-4 by equation-5, we get-

g’/g=[GM/(R+h)2]×[ R2/GM]

  • g’/g=R2/(R+h)2=R2/[R2(1+h/R)2]

  • g’/g=(1+h/R)2 (equation-6)

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If h<<R, then h/R is small enough as compared to 1 so we can neglect its higher powers. So expanding the above equation-6 through binomial distribution, we get-

g’/g=1- 2h/R => g’=g(1- 2h/R) (equation-7)

Thus from the above equation, we can conclude that the value of g decreases with increase in height h above the surface of earth. That’s why the value of g is lesser at mountains as compared to that on flatlands.

b)Effect of depth- We consider earth to be a homogeneous sphere of mass M and radius R with centre at O. Let g be the acceleration due to gravity at the surface of the earth. Then-

g=GM/R2

Here, we calculate how g changes when a body of mass is kept at a depth of 'd' below the earth's surface.(Fig-3)

Let acceleration due to gravity at the depth d be g’. If is the uniform density of earth, then- M=(4/3)πR3

Therefore- g= G×[(4/3)πR3]/ R2 =4/3πGR (equation-8)

Now we need to find the value of g at a point below d distance of earth’s surface. So, the distance of that point from the centre of earth is (R-d). Of course this sphere is smaller than earth. Let its mass be M’ while its density will be the same as we have taken earth to be a homogeneous sphere.

Therefore, g’=GM’/(R-d)2 where M’=4/3(R-d)3 (equation-9)

Now, dividing equation-9 by equation-8, we get-

g’/g=(R-d)/R=1-d/R =>g’=g(1-d/R) (equation-10)

Thus, from the above equation we conclude that acceleration due to gravity decreases with depth.

From (equation-10) we can find the value of g at the centre of the earth. That is-

At centre R=d, then g’=1-R/R=0

Also Read:

Thus at centre of earth value of g is zero

c)Effect of shape of earth- In both the above cases we have taken earth to be a sphere but in reality it is not a perfect sphere. Its shape is an Oblate spheroid. It bulges out at the equator and flattens at the poles.It is an ellipsoid or more specifically a shape of oblate spheroid.(Fig-5)

Hence, the equatorial radius ‘a’ is about 21km larger than the polar radius ‘b’.

Since, g=GM/R2 => g is proportional to 1/R2 as G and M are constants.

And a>b, therefore the value of g is least at equator and keeps increasing towards poles and becomes maximum at poles.

The value of g also varies with rotation of g about its own axis. However we will not cover the derivation in this article.

Also read -

NCERT Physics Notes:

Frequently Asked Questions (FAQs)

1. How will you find the percentage decrease in value of g with height h?

Decrease in value of g is- g-g’=2hg/R.

                Thus fractional decrease in value of g is (g-g’)/g=2h/R

                 Then, percentage decrease in value of g= [(g-g’)/g]×100=[2h/R]×100

2. What will happen if gravity suddenly disappears?

 If gravity suddenly disappears,

  1. all the bodies will be suddenly thrown away from the surface due to earth’s centrifugal  

force.

  1. Motion of the satellites around earth, both natural and artificial will not be possible as no centripetal force will be provided.

  2. All the materials will lose their weights, etc.

3. What are the two factors which determine why some celestial bodies in the solar system have atmosphere and some don’t?

Ability of a celestial body to hold its atmosphere depends on 2 factors mainly- acceleration due to gravity i.e. value of g and its surface temperature.

4. What is the value of g on the moon?

1.625meter per second squared.

5. What is the value of g on the sun?

Using equation-2 and putting values of the sun's mass and radius, the value of g on the sun 

       is found to  be 274.13 meter per second squared.

6. How would the value of 'g' change if the Earth were twice as massive but the same size?
If the Earth were twice as massive but the same size, the value of 'g' would double. This is because 'g' is directly proportional to the mass of the attracting body (in this case, Earth) when the distance from the center remains constant.
7. How does 'g' affect the escape velocity of a planet?
Escape velocity is directly related to 'g' and the radius of the planet. It's calculated as √(2gR), where R is the planet's radius. A higher 'g' value results in a higher escape velocity, making it harder for objects to leave the planet's gravitational influence.
8. Why is the value of 'g' important in rocket science?
The value of 'g' is crucial in rocket science for several reasons: it determines the force needed to escape Earth's gravity, affects fuel consumption calculations, influences trajectory planning, and is used to calculate the rocket's thrust-to-weight ratio, which is critical for lift-off and maneuverability.
9. How does 'g' affect the pressure at the bottom of an ocean?
The value of 'g' directly affects the pressure at the bottom of an ocean. The pressure increases with depth according to the formula P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. A higher 'g' value would result in higher pressure at the same depth.
10. Why do we use 9.8 m/s² as the value of 'g' in many calculations?
We often use 9.8 m/s² as a rounded, convenient approximation of the average value of 'g' on Earth's surface. It simplifies calculations while remaining accurate enough for most practical purposes.
11. What is the difference between 'g' and 'G' in physics?
While 'g' represents the acceleration due to gravity (a local measure), 'G' is the universal gravitational constant. 'G' is used in Newton's law of universal gravitation and has the same value throughout the universe, unlike 'g' which varies depending on location.
12. What is 'g-force' and how is it related to 'g'?
G-force is a measure of acceleration relative to freefall. One 'g' of force is equivalent to the force of gravity at Earth's surface. Higher g-forces, experienced during rapid acceleration or deceleration, are expressed as multiples of 'g' (e.g., 3g means three times the normal force of gravity).
13. What is the relationship between 'g' and the gravitational potential energy?
Gravitational potential energy is directly proportional to 'g'. The formula for gravitational potential energy near Earth's surface is mgh, where m is mass, g is the acceleration due to gravity, and h is height. A higher 'g' value results in more gravitational potential energy for the same mass and height.
14. What is the difference between 'g' and gravitational field strength?
g' and gravitational field strength are essentially the same thing, just expressed differently. Gravitational field strength is defined as the force per unit mass, which is equivalent to the acceleration due to gravity. Both are measured in units of m/s². The term 'gravitational field strength' is often used when discussing the concept more generally, while 'g' is commonly used when referring to specific values.
15. What does 'g' represent in physics?
In physics, 'g' represents the acceleration due to gravity. It's the rate at which an object's velocity changes when falling freely under the influence of gravity, without air resistance or other forces.
16. How would the value of 'g' change if you were on the Moon?
On the Moon, the value of 'g' is much lower, approximately 1.62 m/s². This is about 1/6 of Earth's gravity, which is why astronauts can jump much higher on the Moon despite wearing heavy spacesuits.
17. Can 'g' ever be negative?
The magnitude of 'g' is always positive, as it represents an acceleration towards the center of mass. However, in some problem-solving contexts, we might assign it a negative value to indicate direction, depending on our chosen coordinate system.
18. Why do objects of different masses fall at the same rate in a vacuum?
Objects of different masses fall at the same rate in a vacuum because the acceleration due to gravity ('g') is independent of mass. The greater gravitational force on a more massive object is exactly balanced by its greater inertia, resulting in the same acceleration for all objects.
19. How is 'g' related to weight?
Weight is the force exerted on an object due to gravity. It's calculated by multiplying the object's mass by the local acceleration due to gravity (g). So, weight = mass × g. This is why an object's weight would be different on different planets, but its mass remains constant.
20. How does 'g' affect the period of a pendulum?
The period of a pendulum is inversely proportional to the square root of 'g'. This means that a pendulum will swing more slowly in areas with lower 'g' values, such as at higher altitudes or on celestial bodies with weaker gravity.
21. Is the value of 'g' constant everywhere on Earth?
No, the value of 'g' is not constant everywhere on Earth. It varies slightly due to factors such as latitude, altitude, and local geology. However, the variations are small enough that for many practical purposes, we use an average value.
22. Why does 'g' vary with latitude on Earth?
g' varies with latitude because the Earth is not a perfect sphere. It bulges at the equator due to its rotation, making the radius larger there. Since gravity decreases with distance from the center, 'g' is slightly less at the equator than at the poles.
23. How does altitude affect the value of 'g'?
As altitude increases, the value of 'g' decreases. This is because gravity weakens with distance from the Earth's center. The change is approximately 0.003 m/s² for every kilometer of altitude.
24. Why is the value of 'g' important in physics calculations?
The value of 'g' is crucial because it allows us to calculate the gravitational force acting on objects, determine the motion of falling bodies, and solve problems related to projectile motion, pendulums, and orbital mechanics.
25. What is the standard value of 'g' on Earth's surface?
The standard value of 'g' on Earth's surface is approximately 9.81 m/s² or 32.2 ft/s². This is an average value, as 'g' varies slightly across the Earth's surface due to factors like latitude and altitude.
26. What does CGS stand for in relation to 'g'?
CGS stands for Centimeter-Gram-Second, which is a system of units. In the CGS system, the value of 'g' is expressed as approximately 980 cm/s².
27. What does FPS stand for in relation to 'g'?
FPS stands for Foot-Pound-Second, another system of units. In the FPS system, the value of 'g' is expressed as approximately 32.2 ft/s².
28. How does the value of 'g' differ between CGS and FPS systems?
The numerical value of 'g' appears different in CGS and FPS systems due to the different units used. In CGS, g ≈ 980 cm/s², while in FPS, g ≈ 32.2 ft/s². However, these represent the same physical acceleration, just expressed in different units.
29. How does air resistance affect an object's acceleration compared to 'g'?
Air resistance opposes the motion of a falling object, reducing its acceleration. As a result, the actual acceleration of an object falling through air is less than 'g'. This effect becomes more pronounced as the object falls faster, eventually leading to terminal velocity.
30. What is terminal velocity and how is it related to 'g'?
Terminal velocity is the constant speed reached by an object falling through a fluid (like air) when the drag force equals the gravitational force. While 'g' determines the initial acceleration, terminal velocity is reached when the upward drag force balances the downward force of gravity (mg).
31. How does 'g' affect the boiling point of water?
The value of 'g' indirectly affects the boiling point of water by influencing atmospheric pressure. At higher altitudes where 'g' is slightly lower, atmospheric pressure is reduced, causing water to boil at a lower temperature. However, this effect is primarily due to the change in pressure rather than the change in 'g' itself.
32. Why don't we feel the constant acceleration of 'g'?
We don't feel the constant acceleration of 'g' because we're in equilibrium with it. Our bodies and the surfaces we stand on provide an equal and opposite force (normal force) that counteracts gravity, resulting in no net acceleration. We only feel changes in acceleration, not constant acceleration.
33. How does 'g' affect the speed of sound?
The value of 'g' doesn't directly affect the speed of sound. Sound speed in air primarily depends on temperature, humidity, and air composition. However, 'g' indirectly influences these factors by affecting atmospheric pressure and density at different altitudes.
34. Why is the value of 'g' important in geophysics?
In geophysics, variations in 'g' can provide information about the Earth's internal structure and composition. Gravity surveys measuring small changes in 'g' can help detect underground mineral deposits, map subsurface structures, and contribute to our understanding of plate tectonics.
35. How does 'g' affect the speed of a river?
The value of 'g' influences river speed indirectly. While gravity is the primary force driving river flow (as water moves from higher to lower elevations), the actual speed depends on many factors including slope, channel shape, and water volume. A higher 'g' would increase the potential energy of water at higher elevations, potentially leading to faster flow rates, all other factors being equal.
36. How does 'g' affect the formation of raindrops?
The value of 'g' influences raindrop formation and size. As water droplets fall, they accelerate due to gravity. This acceleration affects the balance between surface tension (which holds the drop together) and air resistance (which can break it apart). In environments with lower 'g', raindrops could potentially grow larger before breaking up.
37. Why is understanding 'g' important in aircraft design?
Understanding 'g' is crucial in aircraft design for several reasons: it affects lift calculations, influences fuel consumption estimates, impacts structural stress analysis (especially for high-g maneuvers), and is essential for determining safe operating parameters and flight envelopes.
38. How does 'g' relate to the concept of weightlessness in space?
Weightlessness in space occurs when an object is in free fall, continuously accelerating due to gravity but feeling no supporting force. This doesn't mean gravity is absent; rather, the object and its surroundings are falling at the same rate. Astronauts in orbit experience weightlessness because they and their spacecraft are in a constant state of free fall around Earth, despite still being influenced by 'g'.
39. Can the value of 'g' ever be zero?
In practice, 'g' is never exactly zero anywhere in the universe due to the presence of mass. However, it can approach zero in certain situations, such as at the center of the Earth (where gravitational forces cancel out), at Lagrange points between celestial bodies, or in deep space far from any significant mass.
40. How does 'g' affect the height of ocean tides?
While 'g' doesn't directly cause tides (they're primarily caused by the gravitational pull of the Moon and Sun), it does influence their height. A stronger 'g' would result in a stronger gravitational interaction between Earth and the Moon/Sun, potentially leading to higher tides. However, tidal heights are also influenced by many other factors like ocean depth and coastline shape.
41. Why is 'g' important in calculating orbital periods?
The value of 'g' is crucial in calculating orbital periods because it determines the strength of the gravitational force keeping an object in orbit. The orbital period is related to the orbital radius and the central body's mass (which determines its 'g' at that radius). Understanding 'g' helps in predicting satellite orbits, planning space missions, and studying celestial mechanics.
42. How does 'g' affect the speed of a falling object in its first second of fall?
Assuming no air resistance, an object falling from rest will reach a speed of approximately 'g' after one second of fall, regardless of its mass. For example, on Earth's surface (g ≈ 9.8 m/s²), an object will be traveling at about 9.8 m/s after falling for one second.
43. What role does 'g' play in determining atmospheric pressure?
The value of 'g' directly influences atmospheric pressure. Atmospheric pressure is the weight of the air column above a given point, and weight is determined by mass and 'g'. The formula for pressure change with altitude involves 'g', explaining why atmospheric pressure decreases with increasing altitude.
44. How would a change in Earth's rotation rate affect 'g'?
If Earth's rotation rate changed, it would affect 'g', particularly at the equator. A faster rotation would increase the centrifugal effect, slightly counteracting gravity and reducing 'g' at the equator. Conversely, a slower rotation would increase 'g' at the equator. The effect would be less pronounced at higher latitudes.
45. Why is 'g' important in studying planetary formation and evolution?
The value of 'g' on a planet or moon provides crucial information about its mass, density, and internal structure. By studying 'g' on different celestial bodies, scientists can infer their composition, understand their formation processes, and track their evolutionary history. This is vital in fields like planetary science and astrobiology.
46. How does 'g' affect the speed of continental drift?
While 'g' doesn't directly drive continental drift, it plays a role in the processes involved. The value of 'g' influences the weight of tectonic plates, affecting the forces involved in subduction and mountain building. It also impacts mantle convection, which is a driving force behind plate tectonics. However, the speed of continental drift is primarily determined by other factors like mantle viscosity and plate boundaries.
47. What is the relationship between 'g' and the shape of the Earth?
The Earth's shape and its 'g' value are intimately related. The Earth's rotation causes it to bulge at the equator, resulting in a slightly oblate spheroid shape. This shape, in turn, causes 'g' to vary with latitude, being slightly less at the equator than at the poles. The study of this relationship falls under the field of geodesy.
48. How does 'g' affect the formation of stalactites and stalagmites?
The value of 'g' influences the formation of stalactites and stalagmites by affecting the rate at which water droplets fall and the force with which they impact surfaces. It also plays a role in the capillary action that draws water through tiny cracks in cave ceilings. While other factors like mineral content and evaporation rate are more significant, 'g' does have a subtle influence on these geological formations.
49. Why is understanding 'g' important in sports science?
Understanding 'g' is crucial in sports science for several reasons: it affects the trajectory of projectiles (like balls or javelins), influences the forces experienced during impacts and landings, and plays a role in determining the energy expenditure of athletes. Knowledge of 'g' is essential for analyzing performance, designing equipment, and understanding the biomechanics of various sports activities.
50. How does 'g' relate to the concept of apparent weight in an elevator?
The concept of apparent weight in an elevator is directly related to 'g'. When an elevator accelerates upward, the apparent weight increases because the acceleration adds to 'g'. Conversely, when the elevator accelerates downward, the apparent weight decreases. If the elevator were to fall freely (accelerating at 'g'), passengers would experience weightlessness, despite still being influenced by gravity.
51. What role does 'g' play in the water cycle?
The value of 'g' plays a significant role in the water cycle. It influences the rate at which raindrops fall, affects the flow of rivers and streams, and plays a part in the process of groundwater percolation. Gravity's influence on water is a key driver in the movement of water through the various stages of the water cycle, from precipitation to surface runoff and groundwater flow.
52. How does 'g' affect the process of sedimentation in bodies of water?
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